Question

x+sqrt(x)=20

Answer 1

\[x+\sqrt{x}=20\] I assume solve for x?

Answer 2

\[(x+\sqrt{x})=20\]=
\[(\sqrt{x}+1)(\sqrt{x})=20\]= (assuming x is positive)
\[\sqrt{x}=4\]=
\[x=16\]
You get this because 16 is the value for which the graph of \[y=x+\sqrt{x}\] and y=20 intersect.

Answer 3

Alternatively:
\[
\begin{align}
x+\sqrt{x}&=20\implies\\
(x+\sqrt x)^2&=400=x^2+2x\sqrt{x}+x
\end{align}
\]Note that:
\[
\begin{align}
x+\sqrt{x}&=20\implies\\
x\sqrt{x}+x^2&=20x
\end{align}
\]Thus we subtract the two equations
\[
\begin{align}
x^2+2x\sqrt{x}+x&=400\\
-\;\;\;\;\;\;2(x^2+x\sqrt{x})&=40x
\end{align}
\]This gives:
\[
-x^2+x=-40x+400
\]This is a quadratic equation with solutions:
\[
x=16, 25
\]

Answer 4

Oh, wow, I'm stupid. A much easier solution:

Let \(x=l^2\), then the equation becomes quadratic:
\[
l^2+l=20
\]
It is obvious that \(l=4, 5\), the, by definition \(x=16, 25\).